Data was randomly collected from plots of soil that contained different fertaliz

Question

Data was randomly collected from plots of soil that contained different fertalizer. The heights of the plants were recorded to test if the mean heights were similar or different for each fertalizer type. Test the hypothesis that the heights are similar for each group using a .05 level of significance: Let mu_1 be the mean of the organic plants, mu_2 be the mean of the non-organic plants, and mu_3 be the mean of the no fertilizer plants. State the hypothesis: Ho: = = & Ha: Compute your test statistic and p-value: F = & p-value: State your conclusion: There (is/is not) statistically significant evidence at the .05 level that the height for the three fertilizer types is .

Explanation / Answer

(The given problem involves one-way ANOVA)

According to the question it is given that

µ1-The mean height of the organic plants

µ2-The mean height of the non-organic plants

µ3-The mean height of the no fertilizer plant

The hypothesis to be tested:

Ho: µ1=µ2=µ3; i.e the mean height of the plants from the different plots of soil is the same.

Against the alternative hypothesis

Ha:µ1µ2µ3; i.e the mean height of the plants from the different plots of soil is not same(atleast a pair of means differ).

The test statistic used is:

F=(SSB/(k-1))/(SSR/(n-k)) ~Fk-1,n-k

where

SSB/(k-1) -Between treatments mean square

SSR/(n-k)-Residual mean square

k- is the number of treatment i.e 3 for the given question

n- is the total number of observations i.e 21 for the given question

The level of significance for the given test is: 0.05

Now we calculate the sum the sum of squares ie SSB and SSR. The formula used for this is as follows:

SST=ki=1 nij=1 (yij-Y)2          (SST is the total sum of squares)

SSB=ki=1 ni(YI –Y)2               (nI is the number of observation i the ith treatment)

SSR=SST -SSB

Using the above formulas we get the following values

SST=9.8895

SSB=2.43577

SSR=9.8895-2.43577=7.45373

Then the value of the test statistic is(put the above values in the formula for the test statistic): F=2.941

and the p-value is: p=0.06104

CONCLUSION

We know (from tables) that the critical value for a F3,18 statistic at 0.05 level of significance is3.160 but here the value of our test statistic is 2.941.. So we conclude that the there is so significant difference in the heights of the plants from the different plots of soil.

Also, the p-value is 0.06104 <0.05. So the test is not significant.

So we accept the null hypothesis.

Thus, there is not statistically significant evidence at the 0.05 level that the mean height for the three fertilizer types is different.

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